Transcript ppt - kaliasgoldmedal

Three Phase Controlled
Rectifiers
Power Electronics by Prof. M. Madhusudhan Rao
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3 Phase Controlled Rectifiers
•
•
•
•
•
Operate from 3 phase ac supply voltage.
They provide higher dc output voltage.
Higher dc output power.
Higher output voltage ripple frequency.
Filtering requirements are simplified for
smoothing out load voltage and load current.
Power Electronics by Prof. M. Madhusudhan Rao
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• Extensively used in high power variable speed
industrial dc drives.
• Three single phase half-wave converters can be
connected together to form a three phase halfwave converter.
Power Electronics by Prof. M. Madhusudhan Rao
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3-Phase
Half Wave Converter
(3-Pulse Converter)
with
RL Load
Continuous & Constant
Load Current Operation
Power Electronics by Prof. M. Madhusudhan Rao
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Power Electronics by Prof. M. Madhusudhan Rao
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Vector Diagram of
3 Phase Supply Voltages
VCN
0
120
120
VAN
0
0
120
VBN
Power Electronics by Prof. M. Madhusudhan Rao
vRN  vAN
vYN  vBN
vBN  vCN
6
3 Phase Supply Voltage Equations
We deifine three line to neutral voltages
(3 phase voltages) as follows
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vRN  van  Vm sin  t ;
Vm  Max. Phase Voltage
vYN
2 

 vbn  Vm sin   t 

3 

 Vm sin  t  120
vBN
0

2 

 vcn  Vm sin   t 

3 

 Vm sin  t  120
0

sin  t  240 
 Vm
0
Power Electronics by Prof. M. Madhusudhan Rao
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van
vbn
Power Electronics by Prof. M. Madhusudhan Rao
vcn
van
9
Each thyristor conducts for 2/3 (1200)
Constant Load
Current
io=Ia
Ia
Ia
Power Electronics by Prof. M. Madhusudhan Rao
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10
To Derive an
Expression for the
Average Output Voltage of a
3-Phase Half Wave Converter
with RL Load
for Continuous Load Current
Power Electronics by Prof. M. Madhusudhan Rao
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11


T1 is triggered at  t        300   
6

 5

T2 is triggered at  t  
    1500   
 6

 7

T3 is triggered at  t  
     2700   
 6

2
0
Each thytistor conducts for 120 or
radians
3
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If the reference phase voltage is
vRN  van  Vm sin  t , the average or dc output
voltage for continuous load current is calculated
using the equation
 56 

3 

Vdc 
Vm sin  t.d  t  


2 
 6 

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3Vm
Vdc 
2
3Vm
Vdc 
2
3Vm
Vdc 
2
 56 



sin

t
.
d

t




 6 

5
 

6


  cos  t 


 

6


 5



  cos  6     cos  6    





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Note from the trigonometric relationship
cos  A  B    cos A.cos B  sin A.sin B 

 5
  cos  6
3Vm 

Vdc 
2 




 5 
 cos    sin   sin  

6

 


 
 
 cos   .cos    sin   sin   
6
6

0
0



cos
150
cos


sin
150
sin  






3Vm


Vdc 
0
0
2 


cos
30
.cos


sin
30
sin











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0
0
0
0



cos
180

30
cos


sin
180

30
sin









3Vm


Vdc 
0
0
2 


cos
30
.cos


sin
30
sin











Note:
cos 1800  300    cos  300 
sin 1800  300   sin  300 
0
0



cos
30
cos


sin
30
sin









3Vm


Vdc 
0
0
2 


cos
30
.cos


sin
30
s
in











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3Vm
 2 cos  300  cos   
Vdc 

2 

3Vm 
3
Vdc 
cos   
2 
2 
2

3Vm
3 3Vm
 3 cos    
Vdc 
cos





2
2
3VLm
Vdc 
cos  
2
Where VLm  3Vm  Max. line to line supply voltage
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The maximum average or dc output voltage is
obtained at a delay angle   0 and is given by
3 3 Vm
Vdc max   Vdm 
2
Where Vm is the peak phase voltage.
And the normalized average output voltage is
Vdcn
Vdc
 Vn 
 cos 
Vdm
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The rms value of output voltage is found by
using the equation
VO RMS 

 3

2

5

6


2
2
Vm sin  t.d  t  




6
1
2
and we obtain
VO RMS 
1

3
 3Vm  
cos 2 
 6 8

Power Electronics by Prof. M. Madhusudhan Rao
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3 Phase Half Wave
Controlled Rectifier Output Voltage
Waveforms For RL Load
at
Different Trigger Angles
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
Van
Vbn
Vcn

V0
=30
0
30
0
60
0
90
0
120

0
0
150
180
0
210
0
240
Van
0
270
0
300
0
330
0
360
Vbn
0
0
390
420
0
60
0
90
0
120
0
0
150
180
0
210
0
240
0
270
t
0
=60
30
=300
Vcn

V0
0
0
0
300
0
Power Electronics by Prof. M. Madhusudhan Rao
330
0
360
0
0
390
420
0
0
=600
t
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
Vbn
Van
Vcn
=900

V0
=90
0
30
0
60
0
90
0
120
0
0
150
180
0
210
0
240
Power Electronics by Prof. M. Madhusudhan Rao
0
270
0
300
0
330
0
360
0
0
390
420
0
0
t
22
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3 Phase Half Wave
Controlled Rectifier With
R Load
and
RL Load with FWD
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T1
T1
a
a
T2
T2
b
b
T3
+
T3
c
c
R V0
R
V0
L
n

n
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3 Phase Half Wave
Controlled Rectifier Output Voltage
Waveforms For R Load
or RL Load with FWD
at
Different Trigger Angles
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Vbn
Van
Vcn
=0
Vs
0
30
0
60
0
90
0
0
120

150
0
0
180
210
0
240
0
270
0
300
0
330
0
360
0
Vbn
Van
0
390
420
0
Vcn
=150
V0
0
30
0
60
0
90
0
0
120
150
0
0
180
210
0
240
0
270
0
300
0
330
Power Electronics by Prof. M. Madhusudhan Rao
0
360
0
0
390
420
0
=00
t
=150
t
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
Vbn
Van
Vcn
=300
V0
0
0
30
0
60
0
90
120
0

0
150
0
0
180 210
0
0
240 270
0
300 330
0
0
360
Vbn
Van
0
0
0
30
0
60
0
90
120
0
0
150
0
0
180 210
0
0
240 270
t
0
390 420
Vcn
=60
V0
=300
0
300 330
Power Electronics by Prof. M. Madhusudhan Rao
0
0
360
0
0
390 420
0
=600
t
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To Derive An
Expression For The Average Or
Dc Output Voltage Of A
3 Phase Half Wave Converter With
Resistive Load
Or
RL Load With FWD
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

T1 is triggered at  t        300   
6

T1 conducts from  30    to 180 ;
0
0
vO  van  Vm sin  t
 5

T2 is triggered at  t  
    1500   
 6

T2 conducts from 150    to 300 ;
0
0
vO  vbn  Vm sin  t  1200 
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 7

0
T3 is triggered at  t  
     270   
 6

T3 conducts from  270    to 420 ;
0
0
vO  vcn  Vm sin  t  240
 Vm
Power Electronics by Prof. M. Madhusudhan Rao

sin  t  120 
0
0
30
30
3
Vdc 
2


  vO .d  t  
 300

1800
vO  van  Vm sin  t ; for  t    30
0
 to 180 
0


  Vm sin  t.d  t  
 300

0
180

3Vm 
Vdc 
  sin  t.d  t  
2  300

3
Vdc 
2
1800
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3Vm
Vdc 
2

  cos  t



  300 

1800
3Vm
0
0


Vdc 

cos180

cos


30



2 
0
cos180  1, we get
3Vm
0


Vdc 
1

cos


30



2 
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Three Phase Semiconverters
• 3 Phase semiconverters are used in Industrial
dc drive applications upto 120kW power
output.
• Single quadrant operation is possible.
• Power factor decreases as the delay angle
increases.
• Power factor is better than that of 3 phase half
wave converter.
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3 Phase
Half Controlled Bridge Converter
(Semi Converter)
with Highly Inductive Load &
Continuous Ripple free Load Current
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Wave forms of 3 Phase Semiconverter
for
 > 600
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3 phase semiconverter output ripple frequency of
output voltage is 3 f S
The delay angle  can be varied from 0 to 
During the period
30   t  210

7
 t 
, thyristor T1 is forward biased
6
6
0
0
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

If thyristor T1 is triggered at  t      ,

6
T1 & D1 conduct together and the line to line voltage
vac appears across the load.
7
, vac becomes negative & FWD Dm conducts.
At  t 
6
The load current continues to flow through FWD Dm ;
T1 and D1 are turned off.
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If FWD Dm is not used the T1 would continue to
conduct until the thyristor T2 is triggered at
 5

 t      , and Free wheeling action would
 6

be accomplished through T1 & D2 .
If the delay angle  

3
, each thyristor conducts
2
for
and the FWD Dm does not conduct.
3
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We deifine three line neutral voltages
(3 phase voltages) as follows
vRN  van  Vm sin  t
;
Vm  Max. Phase Voltage
2 

0
vYN  vbn  Vm sin   t 

V
sin

t

120


 m
3 

2 

0
vBN  vcn  Vm sin   t 
  Vm sin  t  120 
3 

 Vm sin  t  240
0

Vm is the peak phase voltage of a wye-connected source
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vRB  vac   van  vcn  
vYR  vba   vbn  van  
vBY  vcb   vcn  vbn  
vRY  vab   van  vbn  
Power Electronics by Prof. M. Madhusudhan Rao


3Vm sin   t  
6

5 

3Vm sin   t 

6 



3Vm sin   t  
2



3Vm sin   t  
6

43
43
Wave forms of 3 Phase Semiconverter
for
  600
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To derive an
Expression for the
Average Output Voltage
of 3 Phase Semiconverter
for  >  / 3
and Discontinuous Output Voltage
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For  

and discontinuous output voltage:
3
the Average output voltage is found from
3
Vdc 
2
3
Vdc 
2
 7 6



v
.
d

t


ac



 6

7
 6






3
V
sin

t

d

t


m



6



 6

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3 3Vm
Vdc 
1  cos  
2
3VmL
Vdc 
1  cos  
2
VmL  3Vm  Max. value of line-to-line supply voltage
The maximum average output voltage that occurs at
a delay angle of   0 is
Vdc max   Vdm 
3 3Vm

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The normalized average output voltage is
Vdc
Vn 
 0.5 1  cos  
Vdm
The rms output voltage is found from
VO rms 
 3

 2

7

2

v
.
d

t


ac



6 

6
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VO rms 
VO rms 
 3

 2

7


2
2

t

d

t

sin
V
3


m



6



6 

6
 3
 3Vm 
 4
sin 2  


  
2 

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1
2
Average or DC Output Voltage
of a
3-Phase Semiconverter
for  / 3,
and Continuous Output Voltage
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For  

, and continuous output voltage
3
5



6 
2
3 

Vdc 
v
.
d

t

v
.
d

t




ab
ac


2 


2
 6

3 3Vm
Vdc 
1  cos  
2
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Vdc
Vn 
 0.5 1  cos  
Vdm
RMS value of o/p voltage is calculated by using
the equation
VO rms 
VO rms 
 3

 2


2


5
2
vab
.d  t  
6 
 3
 3Vm 
 4
6 

2

vac2 .d  t  


 2

2
 3 cos   

 3

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2
1
2
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Three Phase Full Converter
• 3 Phase Fully Controlled Full Wave Bridge
Converter.
• Known as a 6-pulse converter.
• Used in industrial applications up to 120kW
output power.
• Two quadrant operation is possible.
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• The thyristors are triggered at an interval of
 / 3.
• The frequency of output ripple voltage is 6fS.
• T1 is triggered at t = (/6 + ), T6 is already
conducting when T1 is turned ON.
• During the interval (/6 + ) to (/2 + ),
T1 and T6 conduct together & the output load
voltage is equal to vab = (van – vbn)
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• T2 is triggered at t = (/2 + ), T6 turns off
naturally as it is reverse biased as soon as T2 is
triggered.
• During the interval (/2 + ) to (5/6 + ), T1
and T2 conduct together & the output load
voltage vO = vac = (van – vcn)
• Thyristors are numbered in the order in which
they are triggered.
• The thyristor triggering sequence is 12, 23,
34, 45, 56, 61, 12, 23, 34, ………
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We deifine three line neutral voltages
(3 phase voltages) as follows
vRN  van  Vm sin  t
;
Vm  Max. Phase Voltage
2 

0
vYN  vbn  Vm sin   t 

V
sin

t

120


 m
3 

2 

0
vBN  vcn  Vm sin   t 
  Vm sin  t  120 
3 

 Vm sin  t  240
0

Vm is the peak phase voltage of a wye-connected source
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The corresponding line-to-line
supply voltages are
vRY


 vab   van  vbn   3Vm sin   t  
6



vYB  vbc   vbn  vcn   3Vm sin   t  
2

vBR


 vca   vcn  van   3Vm sin   t  
2

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To Derive An Expression For The
Average Output Voltage Of
3-phase Full Converter
With Highly Inductive Load
Assuming Continuous And
Constant Load Current
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The output load voltage consists of 6 voltage
pulses over a period of 2 radians, Hence the
average output voltage is calculated as

VO dc 
6
 Vdc 
2
2



6
vO .d t ;



vO  vab  3Vm sin   t  
6

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
Vdc 
3

2



6

Vdc 


3Vm sin   t   .d t
6

3 3Vm

cos  
3VmL

cos 
Where VmL  3Vm  Max. line-to-line supply voltage
The maximum average dc output voltage is
obtained for a delay angle   0,
Vdc max   Vdm 
3 3Vm

Power Electronics by Prof. M. Madhusudhan Rao

3VmL

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The normalized average dc output voltage is
Vdcn
Vdc
 Vn 
 cos 
Vdm
The rms value of the output voltage is found from
VO rms 

 6

2



2

2
vO .d  t  




6

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VO rms 
VO rms 
VO rms 

 6

2


 3

2




2
vab .d  t  




6
2


1
2



2
2
3Vm sin   t  .d  t  

6




6
2

1 3 3

 3Vm  
cos 2 
 2 4

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2
1
2
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Three Phase Dual Converters
• For four quadrant operation in many industrial
variable speed dc drives , 3 phase dual
converters are used.
• Used for applications up to 2 mega watt output
power level.
• Dual converter consists of two 3 phase full
converters which are connected in parallel & in
opposite directions across a common load.
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Outputs of Converters 1 & 2
• During the interval (/6 + 1) to (/2 + 1), the
line to line voltage vab appears across the
output of converter 1 and vbc appears across the
output of converter 2
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We deifine three line neutral voltages
(3 phase voltages) as follows
vRN  van  Vm sin  t
;
Vm  Max. Phase Voltage
vYN
vBN
2 

0
 vbn  Vm sin   t 

V
sin

t

120


 m
3 

2 

0
 vcn  Vm sin   t 

V
sin

t

120


 m
3 

 Vm sin  t  240
Power Electronics by Prof. M. Madhusudhan Rao
0

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The corresponding line-to-line
supply voltages are
vRY


 vab   van  vbn   3Vm sin   t  
6



vYB  vbc   vbn  vcn   3Vm sin   t  
2

vBR


 vca   vcn  van   3Vm sin   t  
2

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To obtain an Expression for the
Circulating Current
• If vO1 and vO2 are the output voltages of
converters 1 and 2 respectively, the
instantaneous voltage across the current
limiting inductor during the interval
(/6 + 1)  t  (/2 + 1) is given by
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vr  vO1  vO 2  vab  vbc
 

 

vr  3Vm sin   t    sin   t   
6
2 

 


vr  3Vm cos   t  
6

The circulating current can be calculated by
using the equation
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1
ir  t  
 Lr
t


6
1
ir  t  
 Lr
3Vm
ir  t  
 Lr
ir  max 
1
t


6
vr .d  t 
1


3Vm cos   t   .d  t 
6

 


sin   t  6   sin 1 

 

3Vm

 Lr
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Four Quadrant Operation
Conv. 2
Inverting
2 > 900
Conv. 1
Rectifying
1 < 900
Conv. 2
Rectifying
2 < 900
Conv. 1
Inverting
1 > 900
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• There are two different modes of operation.
 Circulating current free
(non circulating) mode of operation
 Circulating current mode of operation
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Non Circulating
Current Mode Of Operation
• In this mode of operation only one converter is
switched on at a time
• When the converter 1 is switched on,
For 1 < 900 the converter 1 operates in the
Rectification mode
Vdc is positive, Idc is positive and hence the
average load power Pdc is positive.
• Power flows from ac source to the load
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• When the converter 1 is on,
For 1 > 900 the converter 1 operates in the
Inversion mode
Vdc is negative, Idc is positive and the average
load power Pdc is negative.
• Power flows from load circuit to ac source.
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• When the converter 2 is switched on,
For 2 < 900 the converter 2 operates in the
Rectification mode
Vdc is negative, Idc is negative and the average
load power Pdc is positive.
• The output load voltage & load current reverse
when converter 2 is on.
• Power flows from ac source to the load
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• When the converter 2 is switched on,
For 2 > 900 the converter 2 operates in the
Inversion mode
Vdc is positive, Idc is negative and the average
load power Pdc is negative.
• Power flows from load to the ac source.
• Energy is supplied from the load circuit to the
ac supply.
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Circulating Current
Mode Of Operation
• Both the converters are switched on at the same
time.
• One converter operates in the rectification
mode while the other operates in the inversion
mode.
• Trigger angles 1 & 2 are adjusted such that
(1 + 2) = 1800
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• When 1 < 900, converter 1 operates as a
controlled rectifier. 2 is made greater than
900 and converter 2 operates as an Inverter.
• Vdc is positive & Idc is positive and Pdc is
positive.
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• When 2 < 900, converter 2 operates as a
controlled rectifier. 1 is made greater than
900 and converter 1 operates as an Inverter.
• Vdc is negative & Idc is negative and Pdc is
positive.
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